Find all triples $(a,b,c)$ of real numbers for which there exists a function $f:\mathbb{Z}^{+}\rightarrow\mathbb{Z}^{+}$ satisfying $af(n)+bf(n+1)+cf(n+2)<0$ for every $n\in\mathbb{Z}^{+}$ ($\mathbb{Z}^{+}$ denotes the set of positive integers). Proposed by [i]András Imolay[/i], Budapest

Let f:N→N be a function satisfying $ kf(n)\le f(kn)\le kf(n) \plus{} k \minus{} 1$ for all $ k, n\in N$. (a)Prove that $ f(a) \plus{} f(b)\le f(a \plus{} b)\le f(a) \plus{} f(b) \plus{} 1$ for all $ a, b\in N$. (b)If $ f$ satisfies $ f(2007n)\le 2007f(n) \plus{} 200$ for every $ n\in N$, show that there exists $ c\in N$ such that $ f(2007c) \equal{} 2007f(c)$.

Find the function $f(x)$ which satisfies the following integral equation. \[f(x)=\int_0^x t(\sin t-\cos t)dt+\int_0^{\frac{\pi}{2}} e^t f(t)dt\]

$R$ is a ring such that $xy=yx$ for every $x,y\in R$ and if $ab=0$ then $a=0$ or $b=0$. if for every Ideal $I\subset R$ there exist $x_1,x_2,..,x_n$ in $R$ ($n$ is not constant) such that $I=(x_1,x_2,...,x_n)$, prove that every element in $R$ that is not $0$ and it's not a unit, is the product of finite irreducible elements.($\frac{100}{6}$ points)

Find all function(s) $f:\mathbb R\to\mathbb R$ satisfying the equation \[f(x+y)+f(x)f(y)=(1+y)f(x)+(1+x)f(y)+f(xy);\] For all $x,y\in\mathbb R.$

Suppose $a$ and $b$ are two positive real numbers such that the roots of the cubic equation $x^3 - ax + b = 0$ are all real. If $\alpha$ is a root of this cubic with minimal absolute value, prove that \[ \dfrac{b}{a} < \alpha < \dfrac{3b}{2a}. \]

Let be given a parallelogram $ ABCD$ and two points $ A_1$, $ C_1$ on its sides $ AB$, $ BC$, respectively. Lines $ AC_1$ and $ CA_1$ meet at $ P$. Assume that the circumcircles of triangles $ AA_1P$ and $ CC_1P$ intersect at the second point $ Q$ inside triangle $ ACD$. Prove that $ \angle PDA \equal{} \angle QBA$.

Let $\mathbb{N}$ and $\mathbb{N}^*$ be the sets containing the natural numbers/positive integers respectively. We define a binary relation on $\mathbb{N}$ by $a\acute{\in}b$ iff the $a$-th bit in the binary representation of $b$ is $1$. We define a binary relation on $\mathbb{N}^*$ by $a\tilde{\in}b$ iff $b$ is a multiple of the $a$-th prime number $p_a$. i) Prove that there is no bijection $f:\mathbb{N}\to \mathbb{N}^*$ such that $a\acute{\in}b\Leftrightarrow f(a)\tilde{\in}f(b)$. ii) Prove that there is a bijection $g:\mathbb{N}\to \mathbb{N}^*$ such that $(a\acute{\in}b \vee b\acute{\in}a)\Leftrightarrow (g(a)\tilde{\in}g(b) \vee g(b)\tilde{\in}g(a))$.

Do there exist polynomials $a(x)$, $b(x)$, $c(y)$, $d(y)$ such that \[1 + xy + x^2y^2= a(x)c(y) + b(x)d(y)\] holds identically?

Consider the following function. $\textbf{procedure }\textsc{M}(x)$ $\qquad\textbf{if }0\leq x\leq 1$ $\qquad\qquad\textbf{return }x$ $\qquad\textbf{return }\textsc{M}(x^2\bmod 2^{32})$ Let $f:\mathbb N\to\mathbb N$ be defined such that $f(x) = 0$ if $\textsc{M}(x)$ does not terminate, and otherwise $f(x)$ equals the number of calls made to $\textsc{M}$ during the running of $\textsc{M}(x)$, not including the initial call. For example, $f(1) = 0$ and $f(2^{31}) = 1$. Compute the number of ones in the binary expansion of \[ f(0) + f(1) + f(2) + \cdots + f(2^{32} - 1). \]

Find all functions $f$ from positive integers to themselves such that: 1)$f(mn)=f(m)f(n)$ for all positive integers $m, n$ 2)$\{1, 2, ..., n\}=\{f(1), f(2), ... f(n)\}$ is true for infinitely many positive integers $n$.

Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected? $ \textbf{(A)}\ 22 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 27 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 29$

Let $\displaystyle{n \geqslant 3}$ and let $\displaystyle{{x_1},{x_2},...,{x_n}}$ be nonnegative real numbers. Define $\displaystyle{A = \sum\limits_{i = 1}^n {{x_i}} ,B = \sum\limits_{i = 1}^n {x_i^2} ,C = \sum\limits_{i = 1}^n {x_i^3} }$. Prove that: \[\displaystyle{\left( {n + 1} \right){A^2}B + \left( {n - 2} \right){B^2} \geqslant {A^4} + \left( {2n - 2} \right)AC}.\] [i]Proposed by Géza Kós, Eötvös University, Budapest.[/i]

Find all function $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that \[ f(x \plus{} y)(f(x) \minus{} y) \equal{} xf(x) \minus{} yf(y) \] for all $ x,y \in \mathbb{R}$.

Let $r$ be a nonnegative continuous function on the real line. Show that there exists a function $f\in C^1(\mathbb{R})$, not identically zero, such that $f'(x)=f(x-r(f(x)))$, $x\in\mathbb{R}$. (translated by L. Erdős)

Provided that the roots of the polynom $ X^n+a_1X^{n-1} +a_2X^{n-2} +\cdots +a_{n-1}X +a_n:\in\mathbb{R}[X] , $ of degree $ n\ge 2, $ are all real and pairwise distinct, prove that there exists is a neighbourhood $ \mathcal{V} $ of $ \left( a_1,a_2,\ldots ,a_n \right) $ in $ \mathbb{R}^n $ and $ n $ functions $ x_1,x_2,\ldots ,x_n\in\mathcal{C}^{\infty } \left( \mathcal{V} \right) $ whose values at $ \left( a_1,a_2,\ldots ,a_n \right) $ are roots of the mentioned polynom.

Last year, Master Cheung is famous for multi-rotation. This year, he comes to DAMO to make noodles for sweeping monk. One day, software engineer Xiao Li talks with Master Cheung about his job. Xiao Li mainly researches and designs the algorithm to adjust the paramter of different kinds of products. These paramters can normally be obtainly by minimising loss function $f$ on $\mathbb{R}^n$. In the recent project of Xiao Li, this loss function is obtained by other topics. For safety consideration and technique reasons, this topic makes Xiao Li difficult to find the interal details of the function. They only provide a port to calculate the value of $f(\text x)$ for any $\text x\in\mathbb{R}^n$. Therefore, Xiao Li must only use the value of the function to minimise $f$. Also, every times calculating the value of $f$ will use a lot of calculating resources. It is good to know that the dimension $n$ is not very high (around $10$). Also, colleague who provides the function tells Xiao Li to assume $f$ is smooth first. This problem reminds Master Cheung of his antique radio. If you want to hear a programme from the radio, you need to turn the knob of the radio carefully. At the same time, you need to pay attention to the quality of the radio received, until the quality is the best. In this process, no one knows the relationship between the angle of turning the knob and the quality of the radio received. Master Cheung and Xiao Li realizes that minimising $f$ is same as adjusting the machine with multiple knobs: Assume every weight of $\text x$ is controlled by a knob. $f(\text x)$ is a certain performance of the machine. We only need to adjust every knobs again and again and observes the value of $f$ in the same time. Maybe there is hope to find the best $\text x$. As a result, two people suggest an iteration algorithm (named Automated Forward/Backward Tuning, $\text{AFBT}$, to minimise $f$. In $k$-th iteration, the algorithm adjusts the individual weight of $\text{x}_k$ to $2n$ points $\{\text x_k\pm t_k\text e^i:i=1,...,n\}$, where $t_k$ is the step size; then, make $y_k$ be the smallest one among the value of the function of thosse points. Then check if $\text y_k$ sufficiently makes $f$ decrease; then, take $\text x_{k+1}=\text y_k$, then make the step size doubled. Otherwise, make $\text x_{k+1}=\text x_k$ and makes the step size decrease in half. In the algorithm, $\text e^i$ is the $i$-th coordinate vector in $\mathbb{R}^n$. The weight of $i$-th is $1$. Others are $0$; $\mathbf{1}(\cdot)$ is indicator function. If $f(\text x_k)-f(\text y_k)$ is at least the square of $t_k$, then take the value of $\mathbf{1}(f(\text k)-f(y_k)\ge t^2_k)$ as $1$. Otherwise, take it as $0$. $\text{AFBT}$ algorithm Input $\text{x}_0\in \mathbb{R}^n$, $t_0>0$. For $k=0, 1, 2, ...$, perform the following loop: 1: #Calculate loss function. 2: $s_k:=\mathbb{1}[f(\text{x}_k)-f(\text{y}_k)\ge t^2_k]$ #Is it sufficiently decreasing? Yes: $s_k=1$; No: $s_k=0$. 3: $\text{x}_{k+1}:=(1-s_k)\text{x}_k+s_k\text{y}_k$ #Update the point of iteration. 4: $t_{k+1}:=2^{2S_k-1}t_k$ #Update step size. $s_k=1$: Step size doubles; $s_k=0$: Step size decreases by half. Now, we made assumption to the loss function $f:\mathbb{R}^n\to \mathbb{R}$. Assumption 1. Let $f$ be a convex function. For any $\text{x}, \text{y}\in \mathbb{R}^n$ and $\alpha \in [0, 1]$, we have $f((1-\alpha)\text{x}+\text{y})\le (1-\alpha)f(\text{x})+\alpha f(\text{y})$. Assumption 2. $f$ is differentiable on $\mathbb{R}^n$ and $\nabla f$ is L-Lipschitz continuous on $\mathbb{R}^n$. Assumption 3. The level set of $f$ is bounded. For any $\lambda\in\mathbb{R}$, set $\{\text x\in \mathbb{R}^n:f(\text x)\le \lambda\}$ is all bounded. Based on assumption 1 and 2, we can prove that $\left\langle \nabla f(\text x),\text y-\text x \right\rangle \le f(\text y)-f(\text x)\le \left\langle \nabla f(\text x),\text y-\text x\right\rangle+\frac{L}{2}||\text x-\text y||^2$ You can refer to any convex analysis textbook for more properties of convex function. Prove that under the assumption 1-3, for $AFBT$, $\lim_{k \to \infty}f(\text{x}_k)=f^*$

Let $f:[0,1]\rightarrow{\mathbb{R}}$ be a differentiable function, with continuous derivative, and let \[ s_{n}=\sum_{k=1}^{n}f\left( \frac{k}{n}\right) \] Prove that the sequence $(s_{n+1}-s_{n})_{n\in{\mathbb{N}}^{\ast}}$ converges to $\int_{0}^{1}f(x)\mathrm{d}x$.

The function $f$ is given by the table \[\begin{array}{|c||c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 4 & 1 & 3 & 5 & 2 \\ \hline \end{array}\] If $u_0=4$ and $u_{n+1}=f(u_n)$ for $n\geq 0$, find $u_{2002}$. $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

On $[1,2]$ if two functions $f(x)=x^2+px+q$ and $g(x)=x+\frac{1}{x^2}$ get their minumum value at the same point, then the maximum value of $f(x)$ on $[1,2]$ is $\text{(A)}4+\frac{11}{2}\sqrt[3]{2}+\sqrt[3]{4}\qquad\text{(B)}4-\frac{5}{2}\sqrt[3]{2}+\sqrt[3]{4}$ $\text{(C)}1-\frac{1}{2}\sqrt[3]{2}+\sqrt[3]{4}\qquad\text{(D)}$ none above

Let $\mathbb{N} = \{1,2, \ldots\}.$ Does there exists a function $f: \mathbb{N} \mapsto \mathbb{N}$ such that $\forall n \in \mathbb{N},$ $f^{1989}(n) = 2 \cdot n$ ?

Let $ P\subseteq \mathbb{R}^m$ be a non-empty compact convex set and $ f: P\rightarrow \mathbb{R}_{ \plus{} }$ be a concave function. Prove, that for every $ \xi\in \mathbb{R}^m$ \[ \int_{P}\langle \xi,x \rangle f(x)dx\leq \left[\frac {m \plus{} 1}{m \plus{} 2}\sup_{x\in P}{\langle\xi,x\rangle} \plus{} \frac {1}{m \plus{} 2}\inf_{x\in P}{\langle\xi,x\rangle}\right] \cdot\int_{P}f(x)dx.\]

Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ which satisfy \[ f(m + f(n)) = f(m) + n \] for all $m,n \in \mathbb{Z}$.

Let be given $a < b < c$ and $f(x) =\frac{c(x - a)(x - b)}{(c - a)(c - b)}+\frac{a(x - b)(x - c)}{(a - b)(a -c)}+\frac{b(x -c)(x - a)}{(b - c)(b - a)}$. Determine $f(2014)$.

Prove that if $ E \subset \mathbb{R}$ is a bounded set of positive Lebesgue measure, then for every $ u < 1/2$, a point $ x\equal{}x(u)$ can be found so that \[ |(x\minus{}h,x\plus{}h) \cap E| \geq uh\] and \[ |(x\minus{}h,x\plus{}h) \cap (\mathbb{R} \setminus E)| \geq uh\] for all sufficiently small positive values of $ h$. [i]K. I. Koljada[/i]